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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.4.110
Chapter 4, Problem 4.4.110

110. Suppose the derivative of the function y = f(x) is
y'=(x-1)^22(x-2)(x-4).
At what points, if any, does the graph of f have a local minimum, local maximum, or
point of inflection?

Verified step by step guidance
1
Step 1: Identify critical points by setting the derivative y' = (x-1)^2 * 2(x-2)(x-4) equal to zero. Solve for x to find the values where the slope of the tangent line is zero or undefined.
Step 2: Analyze the multiplicity of the factors in y'. For example, (x-1)^2 has even multiplicity, which means it does not change sign at x = 1. Factors with odd multiplicity, such as (x-2) and (x-4), indicate sign changes at x = 2 and x = 4.
Step 3: Use the first derivative test to determine whether each critical point corresponds to a local minimum, local maximum, or neither. Examine the sign of y' on intervals around each critical point (e.g., test values in intervals like (−∞, 1), (1, 2), (2, 4), and (4, ∞)).
Step 4: To find points of inflection, compute the second derivative y'' by differentiating y' = (x-1)^2 * 2(x-2)(x-4). Set y'' equal to zero and solve for x to identify where the concavity changes.
Step 5: Verify the nature of each point of inflection by checking the sign of y'' on intervals around the solutions obtained in Step 4. Points where y'' changes sign indicate points of inflection.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative and Critical Points

The derivative of a function, denoted as y' or f'(x), represents the rate of change of the function at any point. Critical points occur where the derivative is zero or undefined, indicating potential local maxima, minima, or points of inflection. To find these points, we set the derivative equal to zero and solve for x.
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Critical Points

Second Derivative Test

The second derivative test is a method used to determine the nature of critical points. If the second derivative, f''(x), is positive at a critical point, the function has a local minimum; if negative, it has a local maximum. If the second derivative is zero, the test is inconclusive, and further analysis is needed.
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The Second Derivative Test: Finding Local Extrema

Points of Inflection

Points of inflection occur where the concavity of the function changes, which can be identified by analyzing the second derivative. Specifically, a point of inflection exists where f''(x) changes sign, indicating a transition from concave up to concave down or vice versa. These points are important for understanding the overall shape of the graph.
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Related Practice
Textbook Question

35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

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Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


f(x) = x(4 − x)³

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Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

y = 1 / (x² - 1)

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Textbook Question

Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


f'(x) = 2x − 1, P(0,0)

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Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

7. y=sin|x|, -2π≤x≤2π

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Textbook Question

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

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